Use method of reduction of order to find a second solution y2(x) of the homogeneuos equation and a particular solution of the given nonhomogeneous equation:
y" - 4y = 2; y1 = e^-2x
Please use equation and not substitution

The homogeneous equation and a particular solution of the given nonhomogeneous equation is [tex]& y=c_1 e^{-2 x}+c_2 e^{2 x}-\frac{1}{2}[/tex].

The given nonhomogeneous equation is y" - 4y = 2

We are given [tex]$y_1=e^{-2 x}$[/tex]

Let [tex]$y_2=u y_1$[/tex]

Non-homogeneous differential equations are simply differential equations that do not satisfy the conditions for homogeneous equations. In the past, we’ve learned that homogeneous equations are equations that have zero on the right-hand side of the equation.

The two most common methods when finding the particular solution of a non-homogeneous differential equation are:

[tex]$$\begin{aligned}& y_2=u e^{-2 x} \\& y_2^{\prime}=u^{\prime} e^{-2 x}-2 u e^{-2 x} \\& y_2^{\prime \prime}=u^{\prime \prime} e^{-2 x}-2 u^{\prime} e^{-2 x}-2 u^{\prime} e^{-2 x}+4 u e^{-2 x} \\& y_2^{\prime \prime}=u^{\prime \prime} e^{-2 x}-4 u^{\prime} e^{-2 x}+4 u e^{-2 x}\end{aligned}$$[/tex]

Substitute in [tex]$y^{\prime \prime}-4 y=0$[/tex], we get

[tex]$$\begin{aligned}& u^{\prime \prime} e^{-2 x}-4 u^{\prime} e^{-2 x}+4 u e^{-2 x}-4 u e^{-2 x}=0 \\& u^{\prime \prime} e^{-2 x}-4 u^{\prime} e^{-2 x}=0 \\& e^{-2 x}\left(u^{\prime \prime}-4 u^{\prime}\right)=0 \\& u^{\prime \prime}-4 u^{\prime}=0 \\& u^{\prime \prime}=4 u^{\prime}\end{aligned}$$[/tex]

Substitute [tex]$$u^{\prime}=v$ and $u^{\prime \prime}=v^{\prime}$[/tex], we get

[tex]$$\begin{aligned}& v^{\prime}=4 v \\& \frac{d v}{d x}=4 v \\& \frac{1}{v} d v=4 d x\end{aligned}$$[/tex]

Integrate both sides, we get

[tex]$u=\frac{1}{4} C_1 e^{4 x}+C_2$$[/tex]

put in [tex]$y_2=u y_1$[/tex]

[tex]$$y_2=\left(\frac{1}{4} C_1 e^{4 x}+C_2\right) e^{-2 x}$$[/tex]

Choose [tex]$$C_1=4$ and $C_2=0$[/tex], we get

[tex]$$y_2=e^{2 x}$$[/tex]

[tex]$$\begin{aligned}& y_c=c_1 y_1+c_2 y_2 \\& y_c=c_1 e^{-2 x}+c_2 e^{2 x}\end{aligned}$$[/tex]

Particular Solution

[tex]$$\begin{aligned}& y_y=A \\& y_y{ }^{\prime}=0 \\& y_y{ }^{\prime \prime}=0\end{aligned}$$[/tex]

Substitute in [tex]$y^{\prime \prime}-4 y=2$[/tex], we get

[tex]$$\begin{aligned}& 0-4 A=2 \\& -4 A=2 \\& A=-\frac{1}{2} \\& \therefore y_y=-\frac{1}{2}\end{aligned}$$[/tex]

General Solution

[tex]$$\begin{aligned}& y=y_6+y_y \\& y=c_1 e^{-2 x}+c_2 e^{2 x}-\frac{1}{2}\end{aligned}$$[/tex]

Therefore, the second equation is [tex]& y_{2} =c_1 e^{-2 x}+c_2 e^{2 x}-\frac{1}{2}[/tex].

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